8.30

Let $x,y\in R$ and $n\in N$. If $x,y\geq 0$ prove that $x^n=y^n$ if and only if $x=y$

1.$y^n-x^n=(y-x)(y^{n-1}+y^{n-2}x+...+yx^{n-2}+x^{n-1})$

2.use contradiction

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Theorem ($a,b\in R$)

If $a>0$ then $a^{-1}>0$

Proof

Suppose that $a>0$ and $a^{-1}<0$ then $1=aa^{-1}<0\cdot a^{-1}=0$ Contradiction
If $0<a<b$ then $a^{-1}>b^{-1}$

Proof

Since $0<a<b$ then $b-a>0$ and by (1) we have that $a^{-1}-b^{-1}=a^{-1}(b-a)b^{-1}>0$

and thus $a^{-1}>b^{-1}$

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Let $a,c\in R$

If $c\geq 0$ then $|a|\leq c$, if and only if, $-c\leq a\leq c$

Proof

$\Rightarrow$) Suppose that $|a|\leq c$. If $a\geq 0$ then $a=|a|\leq c$ and thus $-c\leq -a\leq a$, therefore $-c\leq a\leq c$.

If $a<0$ then$-a=|a|\leq c$, hence $-c\leq a\leq |a|\leq c$, therefore $-c\leq a\leq c$

$\Leftarrow$) Suppose that $-c\leq a\leq c$ then $-c\leq a$ and $a\leq c$ hence $-a\leq c$ and $a\leq c$, therefore $|a|\leq c$
If $c\geq 0$ then $|a|\geq c$, if and only if, $a\leq-c ~or~a\geq c$

Proof

$\Rightarrow$) Suppose that $|a|\geq c$. If $a\geq 0$ then $a=|a|\geq c$. If $a<0$ then $-a=|a|\geq c$, hence $a\leq -c$

$\Leftarrow$) Suppose that $a\leq -c$ or $a\geq c$. If $a\leq -c$ then $-a\geq c\geq 0$ and thus $|a|=|-a|=-a\geq c$

If $a\geq c\geq 0$ then $|a|=a\geq c$

If $X$ and $Y$ are finite sets then $X\times Y$ is finite and $|X\times Y|=|X||Y|$

Proof

If $x\in X$ note that ${x}\times Y=\{(x,y):y\in Y\}$ is a finite set since Y is finite and $|\{x\}\times Y|=|Y|$

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Now, $\begin{aligned}X\cup Y=\bigcup_{i \in |X|} ({x_i} \times y)\end{aligned} $ , then $X\cup Y$ is finite and $|X\cup Y|=\sum_{i\in|X|}|\{x_i\}\times Y|=\sum_{i\in|X|}|Y|=|X||Y|$

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Prove that the set $Z$ is denumerable

$f(n)=\begin{cases}\frac{n}{2}&\text{ if }n\text{ is even}\\-\frac{n-1}{2}&\text{ if }n\text{ is odd}\end{cases}$

$f:Z\rightarrow N$ is bijective

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